Question

If $a+b=\frac{7}{3}$ and $a^2+b^2=\frac{31}{9},$ find$27(a^3+b^3)$

• A. 154
• B. 156
• C. 152
• D. 164

Answer 1

Answer:

A. 154

Solution:

Given:

$a + b =\frac{7}{3}$

$a^2+b^2=\frac{31}{9}$

Formula used:

(a + b)² = a² + b² + 2ab

a³ + b³ = (a + b)( a² + b² – ab)

Calculation:

According to the question,

⇒ $a + b = \frac{7}{3}$

Squaring both sides,

⇒ $(a+b)^2=(\frac{7}{3})^2$

⇒ $a^2+b^2+2ab=\frac{49}{9}$

⇒ $\frac{31}{9} + 2ab = \frac{49}{9}$

⇒ $2ab = \frac{49}{9}-\frac{31}{9}$

⇒ $2ab = \frac{18}{9}$

⇒ 2ab = 2

⇒ ab = 1

According to the formula,

⇒ a³ + b³ = (a + b)( a² + b² – ab)

Multiply 27 on both sides,

⇒ 27(a³ + b³) = 27(a + b)( a² + b² – ab)

⇒ 27(a³ + b³) = $27(\frac{7}{3})(\frac{31}{9}-1)$

⇒ 27(a³ + b³) = $27(\frac{7}{3})(\frac{22}{9})$

⇒ 27(a³ + b³) = $22\times{7}$

⇒ 27(a³ + b³) = 154

∴ The value of 27(a³ + b³) is 154.